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# C3 Maths A2 AQA 2016 (unofficial mark scheme new)

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Why bother with a post grad course - waste of time? 17-10-2016
1. You need to flip the sign on 5a, you put greater than or equal to not less than or equal to
2. (Original post by Swindelz)
That last italics bit isn't right - you need to sub in the value of x back into the original f(x) first, like matty9
Bloody hell, so I lost the last accuracy mark :')

Ah well, thanks!
3. (Original post by Parhomus)
That's what I had but I also had no clue.
Just realised I crossed out correct working **** my life.
4. (Original post by JamesNuttall)
If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?
What was the actual question
5. (Original post by irMike)
For the stationary point, did you get sin x = 2/3? A few people I spoke to got this, and none of us had any clue how this became a root? Had to keep it as a decimal in the end
If you draw a triangle and label the side opposite angle x as 2 and label the hypotenuse 3 then you can work out the third side by doing root of 3^(2) - 2^(2), which is root 5. So then you could find out the values of tanx and secx using this triangle and yeah I got -root5 as well.
6. (Original post by JamesNuttall)
If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?
****, was this the one where you had to figure out the y co-ordinate afterwards? I crossed out my working for that assuming it was wrong because I had the wrong form... Could've had method marks for it. I deserve the C I'm getting
7. (Original post by matty9)
You need to flip the sign on 5a, you put greater than or equal to not less than or equal to
Fml I only wrote less than and forgot it was the maxima I'm so done with life. Went from 100% to horseshit.
8. (Original post by Parhomus)
Just realised I crossed out correct working **** my life.
Same...
9. Anybody want to join me in committing
10. (Original post by matty9)
I redid question 5a and got the same answer as I did in the exam which was 8ln8-8

Can someone confirm my working?

Attachment 550557
I believe this to be 100% correct
11. (Original post by JamesNuttall)
y = 16x - e^(2x)
dy/dx = 16 - 2e^(2x) which = at a stationary point.
16 = 2e^(2x)
8 = e^(2x)
2x = ln(8)
x = ln(8)/2

d2y/dx2 = -4e^(2x)

when x = ln(8)/2, dy/dx is negetive, therefore it's a max point.

f(x) <= ln(8)/2
Yeah, but you'd have to sub the x you worked out into f(x)
12. Can someone remind me what finding the area of the triangle question was. Coords and all tyvm
13. For the Area of OAB question:

Was it a case of just making x = 0 and y = 0 to find A and B? I'm pretty certain I got A and B correct, then had absolutely no idea what to do to find the area. I tried integrating but I didn't look to me getting anywhere so I moved on because I was running out of time already by that point.
14. ok ive made corrections

(Original post by irMike)
For the Area of OAB question:

Was it a case of just making x = 0 and y = 0 to find A and B? I'm pretty certain I got A and B correct, then had absolutely no idea what to do to find the area. I tried integrating but I didn't look to me getting anywhere so I moved on because I was running out of time already by that point.
you use area of triangle = 1/2ab and bobs ur uncle
15. (Original post by beanigger)
It would help if you could link answers to questions as i cant remember them

Questions:

1) a)Asked to find dy/dx [2]
b) find dy/dx [2]
c) find dy/dx[2]

2) a) show alpha lied between values [2]
b) re-arrange to get into the form x=e(ln5 / x) [3]
c) use iteration thingy where X1 = 2, X3 =2.054 [2]
d) simpsons rule on 5-xx which was 4.49 to 2dp [4]
e) use d) to find approximate for xx which was 1.51 (could someone explain how to do this please) [2]

3) a) solve x2 >|5x-6| *NEED CONFIRMED ANSWERS* [5]

4) a) explain transformation of y=ex onto y=e2x+5 stretch by 1/2 in x-direction and translation by (2.5,0) [4]
b) cant remember

5) a) find range of f(x) = 16x - e2x (differentiate, find stationary point) f(x)>8ln8 - e8ln(8) (i got e8ln(8) but im not sure if it is right) [5]
b) find gg(x) g(x) = 1/x therefore gg(x) = x [2]
6) a) integrate ( ln3x ) / x2 by using by parts twice [4]
b) volume of revolution, integrate ( ( ln3x ) /x2 )2 limits of 1 to 1/3pi( (ln3)2 + 2ln(3) -4 ) [7]

7) a) integrate using substitution (5-2x)(1-4x)1/3 using u=1-4x limits of 1/2 and 1/4 you get 117/224 [7]

8) a) use chain rule to show that (cosx)-1 is secxtanx [2]
b) cant remeber
c) i think it was a 6 marker for the m = 2 and n = 3

9)a) show secx-tanx = -5 goes to secx+tanx=-0.2 [2]
b) find the exact value of cosx cosx=-1/2.6 [3]
c) solve the equation but x is changed into 2x-70 with range of -90 to 90 [3]

i belive i jumbled up a few so pls correct and give CONFIRMED answers, ty.
1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

5a) 8ln8 -8 i think

6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
My final answer was -(ln3)^2 - 2ln3 + 4 for B.

7a) 177/224 not 117/224 (I think you made a typo there.)
16. (Original post by umpa_dumpa)
1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

5a) 8ln8 -8 i think

6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
My final answer was -(ln3)^2 - 2ln3 + 4 for B.

7a) 177/224 not 117/224 (I think you made a typo there.)
Yeah was -2

5a defo correct (see my working)

7a it was either haha have a feeling it was actually 177
17. (Original post by umpa_dumpa)
1 quotient rule. I thin p = -2. (Cant remember if 2 or -2)

For 3a i got x <= 6, 1 <= x <= 2, 3 <= 3

5a) 8ln8 -8 i think

6a) It was only parts once I think, you used parts twice in part 2. In the second part you integrated [ln(3x)]^2/x^2 I think.
My final answer was -(ln3)^2 - 2ln3 + 4 for B.

7a) 177/224 not 117/224 (I think you made a typo there.)
thanks i made the changes xd
18. (Original post by JamesNuttall)
If sin(x) is 2/3, you can draw a right angle triangle to show that the remaining side must be root 5 due to pythagorus, so tan x = 2/root 5 and cos x = root 5 / 3, so sec x is 3/root 5 I think?
How many marks do you reckon one would get if he only made it to sin x=2/3?
19. fak i got like 40 if I'm lucky

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